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The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.?
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Calculation of Relative Lowering of Vapour Pressure
Given data:
- Vapour pressure of pure water at 298K = 3.3kPa
- Mass of glucose = 20g
- Mass of water = 90g
- Molar Mass of glucose = 180g/mol
- Molar Mass of water = 18g/mol
Step 1: Calculate the molality of the solution
- Molality (m) = (moles of solute) / (mass of solvent in kg)
- Mass of glucose in moles = 20g / 180g/mol = 0.111 mol
- Mass of water in kg = 90g / 1000 = 0.09 kg
- Molality (m) = 0.111 mol / 0.09 kg = 1.23 mol/kg
Step 2: Calculate the lowering of vapour pressure
- Relative lowering of vapour pressure (P/P0) = (ΔP/P0) = (Km)
- Kf is the molal depression constant for water which is equal to 0.52 kPa kg/mol.
- P0 is the vapour pressure of the pure solvent.
- ΔP is the lowering in vapour pressure.
- Km is the molal concentration of the solute.
- P0 = 3.3 kPa (given)
- ΔP = P0 x (Km)
- ΔP = 3.3 kPa x 1.23 mol/kg x 0.52 kPa kg/mol
- ΔP = 2.05 kPa
Therefore, the relative lowering of vapour pressure of the solution containing 20g of glucose dissolved in 90g of water at 298K is 2.05 kPa/3.3 kPa = 0.62 or 62%.
Given data:
- Vapour pressure of pure water at 298K = 3.3kPa
- Mass of glucose = 20g
- Mass of water = 90g
- Molar Mass of glucose = 180g/mol
- Molar Mass of water = 18g/mol
Step 1: Calculate the molality of the solution
- Molality (m) = (moles of solute) / (mass of solvent in kg)
- Mass of glucose in moles = 20g / 180g/mol = 0.111 mol
- Mass of water in kg = 90g / 1000 = 0.09 kg
- Molality (m) = 0.111 mol / 0.09 kg = 1.23 mol/kg
Step 2: Calculate the lowering of vapour pressure
- Relative lowering of vapour pressure (P/P0) = (ΔP/P0) = (Km)
- Kf is the molal depression constant for water which is equal to 0.52 kPa kg/mol.
- P0 is the vapour pressure of the pure solvent.
- ΔP is the lowering in vapour pressure.
- Km is the molal concentration of the solute.
- P0 = 3.3 kPa (given)
- ΔP = P0 x (Km)
- ΔP = 3.3 kPa x 1.23 mol/kg x 0.52 kPa kg/mol
- ΔP = 2.05 kPa
Therefore, the relative lowering of vapour pressure of the solution containing 20g of glucose dissolved in 90g of water at 298K is 2.05 kPa/3.3 kPa = 0.62 or 62%.
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The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.?
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The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.?.
The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure water at 298K is 3.3kpa . Calculate the relative lowering of vapour pressure of an aqueous solution containing 20g of glucose dissolved in 90g of water at the same temperature. (Molar Mass of glucose=180gm/mol , Molar Mass of water =18gm/mol.?.
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